# Magnetic Moment of a Charged, Rotating Sphere Problem

1. Apr 13, 2013

### San1405

1. The problem statement, all variables and given/known data

Show for a solid spherical ball of mass m rotating about an axis through its center with a charge q uniformly distributed on the surface of the ball that the magnetic moment μ is related to the angular momentum L by the relation

μ=(5q/6mc)L

2. Relevant equations

μ=IA/c

I sphere =(2/5)m$r^{2}$

3. The attempt at a solution
So I have been at this problem for a few days and haven't gotten very far. I know I need to do a volume integral to account for the total charge of the sphere. But I just got the normal volume of a sphere and that didn't get me the relation. I also think I need to integrate dI with respect to dq and da. But I am not really sure on how to go about that. I am sorry for the lame attempt. I really am just lost. Any help would be really appreciated! Thank you!

Last edited: Apr 13, 2013
2. Apr 13, 2013

### Staff: Mentor

The charge is on the surface only. It is convenient to use some coordinate (angle or z) along the rotation axis, and find dI (current per coordinate) and the area as function of this coordinate. The result follows from a 1-dimensional integral.

3. Apr 13, 2013

### San1405

Thank you so much for the reply!

I am a little confused. I understand that I should be using a dσ. So I get something like

A=∫$r^{2}$sinσdσ.
then
dI=(2/5)m$r^{2}$dσ

This is confusing me because then the pi won't cancel out and I am missing my q. I think I need to add a dq and flip one of my integrals. Am I on the right track?

Thank you again!

4. Apr 13, 2013

### Staff: Mentor

The area needs a factor of pi, and I have no idea why you put 2/5m in the equation for the current. This is independent of the mass, and the interior.

5. Apr 13, 2013

### San1405

Ok, so for the area, I am getting ∏$r^{2}$ after doing the integral.

as for dI, I put the (2/5)m in because that is what I of a sphere equals but then I realized I am looking for dI.
Thus,
dI=∫$r^{2}$dσ
and
I=∫(2/5)m$r^{2}$dI

I am still not sure of the q. Would I be able to use q/(2r)dr=dq and put that I to the equstion of dI=$r^{2}$dq? I tried it but it got me nowhere too....

Sorry for the trouble.

6. Apr 14, 2013

### Staff: Mentor

I think you are mixing two different things here.

- the moment of inertia, coming from the mass. You already have the formula there, you don't need to calculate it. This has nothing to do with the current.
- the current, coming from the charge on the surface. This has nothing to do with the mass.
Both happen to have I as symbol, which is a bit confusing.

Can you calculate the surface charge density?